Difference between revisions of "2022 AMC 10A Problems/Problem 2"

Revision as of 14:28, 15 November 2022

Problem

Mike cycled $15$ laps in $57$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $27$ minutes?

$\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13$

Solution 1

Mike's speed is $\frac{15}{57}=\frac{5}{19}$ laps per minute.

In the first $27$ minutes, he completed approximately $\frac{5}{19}\cdot27\approx\frac{1}{4}\cdot28=\boxed{\textbf{(B) } 7}$ laps.

~MRENTHUSIASM

Solution 2

Mike runs $1$ lap in $\frac{57}{15}=\frac{19}{5}$ minutes. So, in $27$ minutes, Mike ran about $\frac{27}{\frac{19}{5}} \approx \boxed{\textbf{(B) }7}$ laps.

~MrThinker

Solution 3

Mike's rate is $\frac{15}{57}=\frac{x}{27},$ where $x$ is the number of laps he can complete in $27$ minutes. If you cross multiply, $57x = 405$ .

So, $x = \frac{405}{57} \approx \boxed{\textbf{(B) }7}$ .

~Shiloh000

Video Solution 1 (Quick and Easy)

https://youtu.be/tu4rE1nqY9g

~Education, the Study of Everything

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 19: / Line 19: @@
 == Solution 3==
-Mike's rate is <cmath>\frac{15}{57}=\frac{x}{27},</cmath> where <math>x</math> is how many laps he can complete in <math>27</math> minutes.
+Mike's rate is <cmath>\frac{15}{57}=\frac{x}{27},</cmath> where <math>x</math> is the number of laps he can complete in <math>27</math> minutes.
 If you cross multiply, <math>57x = 405</math>.
-So, <math>x = \frac{405}{57} \approx \boxed{\textbf{(B) }7}</math> laps.
+So, <math>x = \frac{405}{57} \approx \boxed{\textbf{(B) }7}</math>.
 ~Shiloh000

Page

Toolbox

Search